CONCRETE STRUCTURES
CONCRETE STRUCTURES 
CONCRETE
BEAM DESIGN
PROBLEM:
A simple supported beam with
a 24 ft span, carries a uniform
factored load of 2.25 kips/ft.
The beam will be exposed to
weather. The specifications call
for Rail, A616-79 Grade 50
rebar. The maximum reinforcement ratio shall be 0.0204. Only 3000 psi concrete
is available. Consider flexure
only and design a rectangular cross section according to ACI
Code latest edition.
I. PROBLEM ANALYSIS:
a. A simple supported beam
b. with a 24 ft span
c. carries a uniform
d. factored load of 2.25 kips/ft
e. The beam will be exposed
to weather.
f. The specifications call for
Rail, A616-79
Grade 50 Rebar.
I. PROBLEM ANALYSIS CONTINUED:
g. The maximum reinforcement
ratio shall be 0.0204
h. Only 3000 psi concrete is
available.
i. Consider flexure only
and design
j. a rectangular cross section
k. according to ACI Code
latest edition
II. ASKED:
a. b = width of beam
b. h = height of beam
c. rebar sizes
III. GIVEN:
a. length of beam
l = 24 ft
b. factored load per unit length
or per unit area
U = 1.4 D + 1.7 L
U = required strength
D = Design Load effect
L = Live Load effect
wu = 2.25 kips/ft
c. Exterior exposure
impacts required
min. concrete coverage
& max. allow.crack width
d. Yield point of non-prestressed reinforcement
fy = 50 kips/sq.in
e. maximum reinforcement ratio 3/4 pb = 0.0204
f. 28-day compression strength of concrete
fc' = 3 kips/sq.in
IV. DIAGRAM
A. ASKED
A rectangular beam b x h
with steel reinforcement
b
h
IV. DIAGRAM
B. GIVEN
A simple beam rests on a support at
each end, the ends are being free
to rotate.
V. Step by Step Solution:
Show all work.
With units of measure.
Maximum Moment at Midspan:
simple beam
uniform linear load
factored load
see formula sheet
figure 5.22, page 118 Parker & Ambrose
M = W * l
----------
8
W = wu * l
Mu = wu * l * l
--------------
8
Mu = 2.25 kips/ft * 24 ft * 24 ft
---------------------------------
8
Mu= 1296 ft kips
-------------
8
Mu = 162 ft kips
============
Minimum Depth hmin:
The minimum depth to limit deflection
is calculated according to Table 3.2
page 76 Leet Book.
Minimum thickness h = l/x
Minimum thickness h
Simply Supported Beams l/16
or ribbed one-way slabs
Adjust h min according fy and
concrete weight see formulas
below table 3.2 page 76 Leet
l/16 for normal weight concrete
& fy = 60,000 lb/sqin
For reinforcement having a yield
point other than 60,000 lb/sqin
multiply the table values by
(0.4 + fy )
----------------------
100,000
hmin = l (0.4 + fy )
------------------------------
16 100,000
hmin = 24 * 12 (0.4 + 500,000)
------------------------------
16 100,000
hmin = 288 (0.4 + 0.5 )
----------------------
16
hmin = 18 * 0.9
hmin = 16.2 in
==========
hmin = minimum height to limit
deflection
this is not the actual height h !
. Calculate bd:
Eq. 3.56 from Leet book p. 104
= (phi) Stress reduction factor
also called Capacity reduction factor
Table 1.4 page 10 Leet
= 0.9 for flexure, axial tension
and combination of
flexure & tension
p = 0.018 selected by student
0.012 would be more on the safe side
solve equation for bd:
Units of Measurement
Insert known values into formula
Mathematically solve formula
The number of steps required
will vary depending on your
calculator and math skills.
You don't need an scientific calculator
Your divider may vary depending
when and if you rounded
0.666, 0.667 or 0.668
Your end result may vary based on
rounding errors.
============
2914.3 based on scientific calculator
2914.5 based on basic 4 function calc
Determine b & d:
d = 15.6 in
========
For short spans the ratio of d/b will
usually run between 1.5 and 2.
Determine h (overall height of beam):
h = d + coverage + radius of rebar
Minimum required concrete coverage
see Table 3.3 Page 105 Leet
Formed Concrete No 6 - No 18 2 inch
exposed to earth No 5 1.5 inch
or weather
Minimum concrete coverage 2 inches.
=======
For preliminary calculations
assume a rebar diameter
( of 1 inch.
r = 0.5 inch
h = d + coverage + radius of rebar
h = 15.6 + 2 + 0.5
h = 18.1 in
round off to the nearest inch
select h = 18 inches
18 inches will be ok, even so it is
0.1 inches smaller than calculated.
h = 18 in > h min 16.2 in from
OK (
fulfills h min. requirement
large beams round off to nearest
even inch 18, 20, 22 etc.
h depth
radius
coverage
Try a second possible configuration:
From step
d = 17.1 in
=========
Determine h (overall height of beam):
h = d + coverage + radius of rebar
Minimum required concrete coverage
see Table 3.3 Page 105 Leet
Formed Concrete No 6 - No 18 2 inch
exposed to earth No 5 1.5 inch
or weather
Minimum concrete coverage 2 inches.
=======
For preliminary calculations
assume a rebar diameter
of 1 inch.
r = 0.5 inch
h = d + coverage + radius of rebar
h = 17.1 + 2 + 0.5
h = 19.6 in
round off to the nearest inch
select h = 20 inches
h = 20 in > h min. 16.2 in OK
fulfills h min. requirement
h depth
radius
coverage
Although the deeper beam is more economical
20 in x 10 in = 200 sq in
12 in x 18 in = 216 sq in
We will use b=12 in & h=18 in to maximize headroom.
Steel required for reinforcement:
Select Rebar Table 7.2 page 38 Leet
# 10 1.27 sq in
# 10 1.27 sq in
# 9 1.00 sq in
Total 3.54 sq in
As actual = 3.54 sq in > 3.37 in As required
O. K.
Actual height h of beam:
If we add
h = d + radius rebar + concrete coverage
#10 rebar = 10/8in
concrete coverage = 2 in per
h = 15.6 in + 0.5/8 in + 2 in
h = 15.6 in + 0.625in + 2 in
0.635 in per book p.38
h = 18.225 in
h = 18.225in > 18 in
to maintain a 2 in concrete coverage
the d has to be reduced by
18.225 - 18 = 0.225
d = 15.6 in - 0.225 in
d = 15.375 in
Although a depth of 18 in will result in a slightly
smaller d if 2 in of coverage is maintained, the
moment capacity will still be adequate since
As supplied is greater than the As required.
Normally the designer can neglect small differences of this magnitude.
The built-in safety factor of the design process
will take care of it.
Also
h = 18 > h min = 16.2 in from
OK
. Check z for control of crack width
As a reinforced concrete beam deflects, the tension side of the beam cracks wherever the low tensile strength of the concrete is exceeded.
The maximum crack width the designer should allow depends on exposure conditions:
Exposure Max. Crack Width
seawater or
wet dry cycles 0.006 in (0.15 mm)
= exposed 0.008 in (0.20 mm) max
protected 0.016 in (0.41 mm)
are permitted by ACI code but may not be
acceptable for the architect who desires a
smooth monolithic concrete surface.
for steel with a yield point up to 40 kips/in
use Gergely-Lutz Eq. 3.28 p. 81 Leet
w = maximum width of crack in 1000th/in
For steel with a yield point over 40 kips/in
use modified Gergely-Lutz Eq. 3.29 p. 81 Leet
z = w / 0.091
fs= stress in steel due to service load
fs = 0.6 x fy
fy = yield strength of rebar
fs = 0.6 x 50 kips/sq in
fs = 30 kips/sq in
dc = distance from tension surface to center
of the row of reinforcing bars closest
to the outside surface
2.4 in was assume
2 in + radius 0.635in = 2.635 in
A = effective tension area of concrete divided
by the number of reinforcing bars
A = AREA OF CONCRETE IN TENSION
NUMBER OF BARS
A = (COVERAGE +RADIUS)x 2 x b
As / cross sectional area of bar
A = (2.4 in) (2) (12 in)
3.53 in / 1.27 in
A = 20.7 sq in
from (((
((((
((((
z = 110 kips/in
((((
ACI code (p. 81 Leet) specifies that z not to exceed
exterior exposure 145 kips/in
0.013 in (0.33mm) crack
interior exposure 175 kips/in
0.016 in (0.41mm) crack
((((
z = 110 kips/in <145 kips/in OK
(((( Check Rebar spacing:
((((
Spacing between bars must not be less than
1 inch or 1 bar diameter (.
((((
Allowing 2 inch of side cover on each side.
((((
The available spacing between bars equals.
b - ( side cover (2) + n rebar diameters) = s n - 1
12 - (2 in (2) + 2 (10/8) + 1(9/8) = 2.2 in
2 spaces
((((
s = 2.2 in > 10/8 > 1 inch OK
Shear Reinforcement:
In addition to the flexural steel, U-shaped reinforcement, called stirrups, is provided for shear.
To ensure that the reinforcement will remain
in position when the concrete is placed, a rigid cage is formed by wiring the stirrups to the main flexural steel at the bottom and to small ( diameter bars, used to anchor the stirrups, at the top. The top steel is not designed but arbitrarily selected by the designer.
